Sep 6, 2016 - N(s)=k si...

0 downloads 1 Views 121KB Size

On the q-analogue of the Maurer-Cartan equation Mauricio Angel and Rafael D´ıaz September 6, 2016 Abstract We consider deformations of the differential of a q-differential graded algebra. We prove that it is controlled by a generalized Maurer-Cartan equation. We find explicit formulae for the coefficients ck involved in that equation.

Introduction Following the work of Kapranov [6] the interest on the theory of N-complexes has risen. A delicate issue in the theory is the appropriate definition of a N-differential algebra. At the moment there exists two options for such a definition, each describing very distinct mathematical phenomena. In [2] the notion of N-differential graded algebra A has been defined as follows: A must be a graded associative algebra provided with an operator d : A → A of degree 1 such that d(ab) = d(a)b + (−1)a¯ ad(b) and dN = 0. The study of the deformations of the differential of a N-dga has also been initiated in [2]. Applications to differential geometry are given in [3]. Applications to the study of A∞ -algebras of depth N, are contained in [4]. The other possible N-generalization of the notion of a differential algebra (see [7],[8],[9]) goes as follows: One fixed a primitive N-th root of unity q. One considers q-differential graded algebras A defined by: A must be a graded associative algebra provided with an operator d : A → A of degree 1 such that d(ab) = d(a)b + q a¯ ad(b) and dN = 0. In this paper we consider deformations of the differential of a q-differential graded algebra. We prove that it is controlled by a generalized Maurer-Cartan equation. We find explicit formulae for the coefficients ck involved in that equation.

1

1

q-differential graded algebras and modules

Let k be a commutative ring with unity. For q ∈ k the q-numbers are given by [ ]q : N → k, k 7−→ [k]q = 1 + q + · · · q k−1 and [k]q ! = [1]q [2]q · · · [k]q . We assume that on k, for N ∈ N fixed, we can choose q ∈ k such that [N]q = 0 and [k]q is invertible for 1 ≤ k ≤ N − 1, in this case we refer to q as a primitive N-th root of unity. Definition 1 A q-differential graded algebra or q-dga over k is a triple (A• , m, d) where m : Ak ⊗ Al → Ak+l and d : Ak → Ak+1 are k-modules morphisms satisfying 1) (A• , m) is a graded associative k-algebra. 2) d satisfies the q-Leibniz rule d(ab) = d(a)b + q a¯ ad(b). 3) If q is a primitive N-th root of unity, then dN = 0, i.e., (A• , d) is a N-complex. Definition 2 Let (A• , mA , dA ) be a q-dga and M • be a k-module. A q-differential module (q-dgm) over (A• , mA , dA ) is a triple (M • , mM , dM ) where mM : Ak ⊗ M l → M k+l and dM : M k → M k+1 are k-module morphisms satisfying the following properties 1. mM (a, mM (b, φ)) = mM (mA (a, b), φ), for all a, b ∈ A• and φ ∈ M • . We denote mM (a, φ) by aφ, for all a ∈ A and φ ∈ M. 2. dM (aφ) = dA (a)φ + q a¯ adM (φ) for all a ∈ A• and φ ∈ M • . 3. If q is a primitive N-th root of unity, then dN M = 0.

2

N-curvature

Let (M • , mM , dM ) be a q-dgm over a q-dga (A• , mA , dA ). For a ∈ M 1 we define a deformation of dM as follows for all φ ∈ M • .

Dφ = dM φ + aφ,

In order to obtain a q-dgm structure on (M • , mM , D) we require that D N = 0. In [KN] the consecutive powers of D are calculated and the following formula is obtained N −1 X N N N (D k−1 a)dN −k φ + (D N −1 a)φ, D φ = dM φ + k q k=1 where

N k q

is the q-binomial coefficients given by [N]q ! N . = k q [N − k]q ![k]q ! 2

For q a primitive N-th root of unity the formula above reduces to D N φ = (D N −1 a)φ, because in this case dN M = 0 and 1−q N 1−q

N k q

= 0.

(1)

= 0 for 0 < k < N, since [N]q = 1+q +· · ·+q N −1 =

Formula (1) is the first step towards the construction of the q-analogue of the MaurerCartan equation. A further step is required in order to write down D N −1 a in terms of dM and a only. We shall work with a slightly more general problem: finding an explicit expression for D N φ given in terms of dM and a only. First we review some notations from [AD]. P For s = (s1 , ..., sn ) ∈ Nn we set l(s) = n and |s| = i si . For 1 ≤ i < n, s>i denotes the vector given by s>i = (si+1 , ..., sn ), for 1 < i ≤ n, s

(l)

then ea = dlEnd(ea ) reduces to ea = edl (a) , thus (s1 ) e(s) · · · ea(sn ) = eds1 (a) · · · edsn (a) . a = ea

For N ∈ N we define EN = {s ∈ N(∞) : |s| + l(s) ≤ N} and for s ∈ EN we define N(s) ∈ Z by N(s) = N − |s| − l(s). We introduce a discrete quantum mechanical

1

by

1. V = N(∞) . 1

a discrete quantum mechanical system is given by the following data (1) A directed graph Γ (finite or infinite). (2) A map L : EΓ → R called the Lagrangian map of the system. An associated Hilbert space H = CVΓ . Operators Un : H → H, where n ∈ Z, given by X (Un f )(y) = ωn (y, x)f (x), x∈VΓ

where the discretized kernel ωn (y, x) admits the following representation X Y ωn (y, x) = eiL(e) . γ∈Pn (Γ,x,y) e∈γ

Pn (Γ, x, y) denotes the set of length n paths in Γ from x to y, i.e., sequences (e1 , · · · , en ) of edges in Γ such that s(e1 ) = x, t(ei ) = s(ei+1 ), i = 1, ..., n − 1 and t(en ) = y.

3

2. There is a unique directed edge e from vertex s to t if and only if t ∈ {(0, s), s, (s + ei )} where ei = (0, .., |{z} 1 , .., 0) ∈ Nl(s) , in this case we set source(e) = s and i−th

target(e) = t.

3. Edges are weighted according to the following table source(e) s s s

target(e) (0, s) s (s + ei )

v(e) 1 q |s|+l(s) q |s

The set PN (∅, s) consists of all paths γ = (e1 , ..., eN ), such that source(e1 ) = ∅, target(eN ) = s and source(el+1) = target(el ). For γ ∈ PN (∅, s) we define the weight v(γ) of γ as N Y v(el ). v(γ) = l=1

The following result is proven as Theorem 17 in [AD]. Theorem 3 Let (M • , mM , dM ) be a q-dgm over a q-dga (A• , mA , dA ). For a ∈ M 1 consider the map Dφ = dM φ + aφ. Then we have N

D =

N −1 X

ck dkM

k=0

where ck =

X

cq (s, N)a(s)

and

s∈EN N (s)=k si

cq (s, N) =

X

v(γ).

γ∈PN (∅,s)

Suppose that we have a q-dgm (M • , mM , dM ), q a 3-rd primitive root of unity, and we want to deform it to D = dM + a, a ∈ M 1 . So we required that D 3 = 0. By Theorem 2 X 3 we must have ck dkM = 0. Let us compute the coefficients ck . First, notice that k=0

E3 = {∅, (0), (1), (2), (0, 0), (1, 0), (0, 1), (0, 0, 0)} We proceed to compute the coefficients ck for 0 ≤ k ≤ 2 using Theorem 3. k=0 There are four vectors in E3 such that N(s) = 0, these are (2), (1, 0), (0, 1), (0, 0, 0). 4

For s = (2) the only path from ∅ to (2) of length 3 is ∅ → (0) → (1) → (2) (2) weight 1 and ea = d2M (ea ), thus c(s, 3) = d2M (a).

with

For s = (1, 0) the only path from ∅ to (1, 0) of length 3 is ∅ → (0) → (0, 0) → (1, 0) (1,0) with weight 1 and ea = dM (ea )ea , thus c(s, 3) = dM (a)ea . For s = (0, 1) the paths from ∅ to (0, 1) of length 3 are ∅ → (0) → (0, 0) → (0, 1) (0,1) with weight q; and ∅ → (0) → (1) → (0, 1) with weight 1. Since ea = ea dM (ea ), then c(s, 3) = (1 + q)ea dM (ea ). For s = (0, 0, 0) the only path from ∅ to (0, 0, 0) of length 3 is ∅ → (0) → (0, 0) → (0,0,0) (0, 0, 0) whose weight is 1 and ea = a3 , thus c(s, 3) = a3 . k=1 There are two vectors in E3 such that N(s) = 1, namely (1) and (0, 0). For s = (1), the paths from ∅ to (1) of length 3 are ∅ → ∅ → (0) → (1) with weight 1; ∅ → (0) → (0) → (1) with weight q; ∅ → (0) → (1) → (1) with weight q 2 . (1) ea = dM (ea ) and thus cq (s, 3) = (1 + q + q 2 ) = 0. For s = (0, 0) the paths from ∅ to (0, 0) of length 3 are ∅ → (0) → (0) → (0, 0) with weight q. ∅ → ∅ → (0) → (0, 0) with weight 1. ∅ → (0) → (0, 0) → (0, 0) with weight (0,0) q 2 . ea = a2 and cq (s, 3) = 0. k=2 (0) is the only vector in E3 such that N(s) = 2. The paths from ∅ to (0) of length 3 are ∅ → ∅ → ∅ → (0) with weight 1; ∅ → (0) ∅ → (0) → (0) with weight q; ∅ → (0) → (0) → (0) with weight q 2 . ea = a and c(s, 3) = (1 + q + q 2 ) = 0. So we have proven that the 3-curvature is given by D 3 = d2M (a) + dM (a)a + (1 + q)adM (a). ¯

Notice that in case of q-commutativity αβ = q α¯β βα, the 3-curvature reduces to D 3 = d2M (a). Suppose now that q is a 4-th primitive root of unity and we required that D 4 = 0. 3 X By Theorem 3 we must have ck dkM = 0. Notice that k=0

E4 = {∅, (0), (1), (2), (3), (0, 0), (1, 0), (0, 1), (2, 0), (0, 2), (1, 1), (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1), (0, 0, 0, 0)} 5

We proceed to compute the coefficients ck for 0 ≤ k ≤ 3 using Theorem 3. k=3 j i ∅ −→ ∅ −→ (0) −→ (0)

the weight is

X

q j and c3 = (1 + q + q 2 + q 3 )a = 0.

i+j=3

k=2 j i k ∅ −→ ∅ −→ (0) −→ (0) −→ (0, 0) −→ (0, 0)

the weight is

X

q j q 2k and

X

q j q 2k and

i+j+k=2

cq ((0, 0), 4)a(0,0) = (1 + q 2 )a2 . j

i

k

∅ −→ ∅ −→ (0) −→ (0) −→ (1) −→ (1)

the weight is

i+j+k=2

cq ((1), 4)a(1) = (1 + q 2 )d(a). Finally, c2 = (1 + q 2 )(a2 + d(a)). k=1 j i k l ∅ −→ ∅ −→ (0) −→ X (0) −→ (0, 0) −→ (0, 0) −→ (0, 0, 0) −→ (0, 0, 0) q j q 2k q 3l and cq ((0, 0, 0), 4)a(0,0,0) = (1 + q + q 2 + q 3 )a3 = 0. the weight is i+j+k+l=1

j

i

k

l

∅ −→ ∅ −→ (0) −→ (0) −→ (1) −→ (1) −→ (0, 1) −→ (0, 1) X q j q 2k q 3l . the weight is i+j+k+l=1 i

j

k

l

j

k

l

∅ −→ ∅ −→ (0) −→ (0) −→ (0, 0) −→ (0, 0) −→ (0, 1) −→ (0, 1) X the weight is q j q 2k qq 3l and cq ((0, 1), 4)a(0,1) = ((1 + q + q 2 + q 3 ) + (1 + q + q 2 + i+j+k+l=1

3

q ))ad(a) = 0. i

∅ −→ ∅ −→ (0) −→ (0) −→ (0, 0) −→ (0, 0) −→ (1, 0) −→ (1, 0) X the weight is q j q 2k q 3l and cq ((1, 0), 4)a(1,0) = (1 + q + q 2 + q 3 )d(a)a = 0. i+j+k+l=1 i

the weight is

j

k

l

∅ −→ ∅ −→ (0) −→ (0) −→ (1) −→ (1) −→ (2) −→ (2) X q j q 2k q 3l and cq ((2), 4)a(2) = (1 + q + q 2 + q 3 )d2 (a) = 0.

i+j+k+l=1

Finally, c1 = cq ((0, 0, 0), 4)a(0,0,0) +cq ((0, 1), 4)a(0,1) +cq ((1, 0), 4)a(1,0) +cq ((2), 4)a(2) = 0. k=0 ∅ −→ (0) −→ (0, 0) −→ (0, 0, 0) −→ (0, 0, 0, 0) the weight is 1 and cq ((0, 0, 0, 0), 4)a(0,0,0,0) = a4 . 6

∅ −→ (0) −→ (0, 0) −→ (0, 0, 0) −→ (0, 0, 1) ∅ −→ (0) −→ (1) −→ (0, 1) −→ (0, 0, 1) ∅ −→ (0) −→ (0, 0) −→ (0, 1) −→ (0, 0, 1) the weight is 1 + q + q 2 and cq ((0, 0, 1), 4)a(0,0,1) = (1 + q + q 2 )a2 d(a). ∅ −→ (0) −→ (0, 0) −→ (1, 0) −→ (0, 1, 0) ∅ −→ (0) −→ (0, 0) −→ (0, 0, 0) −→ (0, 1, 0) the weight is 1 + q and cq ((0, 1, 0), 4)a(0,1,0) = (1 + q)ad(a)a. ∅ −→ (0) −→ (0, 0) −→ (0, 0, 0) −→ (1, 0, 0) the weight is 1 and cq ((1, 0, 0), 4)a(1,0,0) = d(a)a2 . ∅ −→ (0) −→ (0, 0) −→ (1, 0) −→ (2, 0) the weight is 1 and cq ((2, 0), 4)a(2,0) = d2 (a)a. ∅ −→ (0) −→ (0, 0) −→ (0, 1) −→ (0, 2) ∅ −→ (0) −→ (1) −→ (0, 1) −→ (0, 2) ∅ −→ (0) −→ (1) −→ (2) −→ (0, 2) the weight is 1 + q + q 2 and cq ((0, 2), 4)a(0,2) = (1 + q + q 2 )ad2 (a). ∅ −→ (0) −→ (0, 0) −→ (1, 0) −→ (1, 1) ∅ −→ (0) −→ (1) −→ (0, 1) −→ (1, 1) the weight is 1 + q 2 and cq ((1, 1), 4)a(1,1) = (d(a))2 . ∅ −→ (0) −→ (1) −→ (2) −→ (3) the weight is 1 and cq ((3), 4)a(3) = d3 (a). Finally, c0 = a4 + (1 + q + q 2 )a2 d(a) + (1 + q)ad(a)a + d(a)a2 + d2 (a)a + (1 + q + q 2 )ad2 (a) + (d(a))2 + d3 (a). So we have that the 4-curvature is given by D 4 = (1 + q 2 )(a2 + d(a))d2M + a4 + (1 + q + q 2 )a2 d(a) + (1 + q)ad(a)a+ d(a)a2 + d2 (a)a + (1 + q + q 2 )ad2 (a) + (d(a))2 + d3 (a). In the remainder of the paper we consider infinitesimal deformations of a q-differential. 7

Theorem 4 Let (M • , mM , dM ) be a q-dgm such that dN M = 0, consider the infinitesimal 1 • 2 deformation D = dM + te, where e ∈ End (M ) and t = 0, then N −1 X X −k−1 q |v|+w(v) dN , DN = M k=0

v∈P ar(N,N −k−1)

where P ar(N, N − k − 1) = {(v0 , · · · , vN −k−1)/ P |v| = v1 + · · · + vn and w(k) = ivi . Proof: By Theorem 3, D N =

PN −1 k=0

X

vi = N},

ck dkM . Since t2 = 0, then

(te)(s) = (te)(s1 ) · · · (te)(sl(s) ) = tl(s) e(s) = 0 unless l(s) ≤ 1. Thus EN is given by EN = {(0), (1), · · · , (N − 1)}. For 0 ≤ k ≤ N −1, since N(s) = N −|s|−l(s) = k and l(s) = 1 we obtain |s| = N −k −1, and the only vector s in EN of length 1 such that |s| = N −k−1 is precisely s = (N −k−1). Thus X cq (s, N)a(s) = cq ((N − k − 1)), N)a(s) = cq ((N − k − 1)), N)dN −k−1(a). ck = s∈EN N (s)=k si

Any path from ∅ to (N − k − 1) of length N must be of the form ∅| → ·{z · · → ∅} → (0) → · · · → (0) → (1) → · · · (1) → · · · (N − k − 1) → · · · → (N − k − 1) {z } {z } {z } | | | p0

p1

p2

pN−k−1

with p0 + p1 + · · · + pN −k−1 = N. The weight of such path is q p1 q p2 (1+1) ...q pN−k−1 (N −k) = q |p|+w(p), where p = (p0 , · · · , pN −k−1). Then X X cq (s, N) = v(γ) = q |p|+w(p). p∈P ar(N,N −k−1)

γ∈PN (∅,s)

Acknowledgement We thank Sylvie Paycha and Nicol´as Andruskiewitsch.

8

References [1] V. Abramov, R. Kerner, On certain realizations of the q-deformed exterior differential calculus, Reports on Math. Phys., 43 (1999) 179-194 [2] M. Angel, R. D´ıaz, N-differential graded algebras, Preprint math.DG/0504398. [3] M. Angel, R. D´ıaz, N-flat connections, Preprint math.DG/0511242. [4] M. Angel, R. D´ıaz, AN ∞ -algebras, In preparation. [5] M. Dubois-Violette, Lectures on differentials, generalized differentials and some examples related to theoretical physics.-Quantum symmetries in theoretical physics and mathematics (Bariloche 2000), pages 59-94, Contemp. Math. 294, A.M.S. 2002. [6] M.M. Kapranov, On the q-analog of homological algebra.-Preprint q-alg/9611005. [7] C. Kassel et M. Wambst, Alg`ebre homologique des N-complexes et homologie de Hochschild aux racines de l’unit´e.-Publ. Res. Inst. Math. Sci. Kyoto University 34 (1998), n 2, 91-114. [8] R. Kerner, B. Niemeyer, Covariant q-differential calculus and its deformations at q N = 1, Lett. in Math. Phys., 45,161-176, (1998). [9] A. Sitarz; On the tensor product construction for q-differential algebras.-Lett. Math. Phys. 44 1998.

Mauricio Angel. Universidad Central de Venezuela (UCV). [email protected] Rafael D´ıaz. Universidad Central de Venezuela (UCV). [email protected]

9